I have the question:

Given that $$y=\frac{e^x}{\sqrt{1+2x}}$$

Show that $$\frac{dy}{dx} = \frac{2xe^x}{\sqrt{(1+2x)^3}}$$

I've done the question but I got $2xe^x(\sqrt{(1+2x)^3})$. I feel like this is too similar to the sheet's answer, so am I wrong or is the sheet printed wrong?


Let $u = e^x$, $v = 1+2x$, $w = \sqrt{v}$.

Using the chain rule, $$\frac{dw}{dx} = \frac{dw}{dv} * \frac{dv}{dx}$$

So $$\frac{dw}{dx} = 2(\frac{1}{2}v^{-\frac{1}{2}}) = \frac{1}{\sqrt{v}} = \frac{1}{\sqrt{1+2x}}$$

And $\frac{du}{dx}$ is clearly $e^x$.

Therefore, using the quotient rule, $$\frac{dy}{dx} = \frac{e^x(\sqrt{1+2x})- e^x(\frac{1}{\sqrt{1+2x}})}{(1+2x)^2}$$




Using the rule

$$\frac{x^m}{x^n} = x^{m-n}$$

We get


$$=2xe^x(\sqrt{1+2x})^3 $$

Have I done this correctly?


3 Answers 3


I suspect

$$\frac{1}{\left(\sqrt{1+2x}\right)^2} \not = \frac{1}{\left({1+2x}\right)^2}$$

and $$\frac{\frac{1}{\sqrt{1+2x}}}{(1+2x)^2} \not =\frac{1}{(1+2x)^{-\frac{3}{2}}}$$

but $$\frac{\frac{1}{\sqrt{1+2x}}}{\left(\sqrt{1+2x}\right)^2} = \frac{1}{\left(\sqrt{1+2x}\right)^3}$$

  • $\begingroup$ Ah, I see now. Thank you for your help :) $\endgroup$
    – imulsion
    Dec 3, 2014 at 21:27

The error is in $$\frac1{\sqrt{2x+1}} \cdot \frac1{(2x+1)^2} = \frac1{(2x+1)^{2+\frac12}} = \frac1{(2x+1)^{\frac52}}$$ and $$\frac1{\sqrt{2x+1}^2} \ne \frac1{(2x+1)^2}$$ so there is an error in the quotient rule. (the exponent is positive in the denominator, not negative)
Tracing the errors through the equation we obtain $$\frac1{\sqrt{2x+1}} \cdot \frac1{(\sqrt{2x+1})^2} = \frac1{\sqrt{2x+1}^3} = \frac1{(2x+1)^{\frac32}}$$


Using Quotient rule: $$\frac{dy}{dx} = \frac{e^x(\sqrt{1+2x})-e^x(\frac{1}{\sqrt{1+2x}})}{1+2x}.$$ You can simplify now.

  • $\begingroup$ I just show that the quotient rule was applied incorrectly in first step. $\endgroup$
    – Alex Silva
    Dec 3, 2014 at 21:30
  • 1
    $\begingroup$ Yes. $(\sqrt{1+2x})^2 = 1+2x$. It was I have written. $\endgroup$
    – Alex Silva
    Dec 3, 2014 at 21:32
  • $\begingroup$ But $v = 1+2x$, not $\sqrt{1+2x}$ $\endgroup$
    – imulsion
    Dec 3, 2014 at 21:33
  • $\begingroup$ I know. But it was you that did this mistake. My expression is correct. I don't understand what are you talking about! $\endgroup$
    – Alex Silva
    Dec 3, 2014 at 21:36
  • $\begingroup$ Actually...yes, you are right. But I still managed to come out with the correct answer... $\endgroup$
    – imulsion
    Dec 3, 2014 at 21:37

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